(\partial)/(\partial y)(4+x^3+y^3-3xy)

解

\frac{\partial}{\partial y} (4 + x^{3} + y^{3} - 3 x y)

3 y^{2} - 3 x

解答ステップ

\frac{\partial}{\partial y} (4 + x^{3} + y^{3} - 3 x y)

x

を定数として扱う

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (4) + \frac{\partial}{\partial y} (x^{3}) + \frac{\partial}{\partial y} (y^{3}) - \frac{\partial}{\partial y} (3 x y)

\frac{\partial}{\partial y} (4) = 0

\frac{\partial}{\partial y} (x^{3}) = 0

\frac{\partial}{\partial y} (y^{3}) = 3 y^{2}

\frac{\partial}{\partial y} (3 x y) = 3 x

= 0 + 0 + 3 y^{2} - 3 x

簡素化

= 3 y^{2} - 3 x